Probability of gaps between coordinates of a random point on the sphere Let $X=(X_1,\ldots,X_n)$ be a point chosen uniformly at random from the sphere $S^{n-1}\subseteq \mathbb R^n$. Given $a>0$, what is the probability that $|X_1|^2-|X_i|^2\geq a$ for all $i>1$? Is there a formula (explicit, asymptotic, etc.) for this probability in terms of $a$ and $n$?
 A: $\newcommand\Z{\mathbf Z}\newcommand\ep\varepsilon\newcommand\tP{\tilde P}\newcommand\de\delta\newcommand\R{\mathbb R}$Note that the random point $\Z/|\Z|$ is uniformly distributed on the sphere $S^{n-1}$, where
$\Z:=(Z_1,\dots,Z_n)$, $|\Z|:=\sqrt{Z_1^2+\dots+Z_n^2}$, and $Z,Z_1,\dots,Z_n$ are iid standard normal random variables.
So, the probability in question is
\begin{equation}
    P:=P(Z_i^2\le Z_1^2-a|\Z|^2\ \forall i\ge2). 
\end{equation}
Let
\begin{equation}
    a=\frac{c\ln n}n,
\end{equation}
where $c$ is a nonnegative real constant; we shall see that this is the right choice for $a$.
Using your favorite exponential inequality, we see that eventually (that is, for all large enough $n$)
\begin{equation}
    P(|\,|\Z|^2-n|>\sqrt n\,\ln n)\le\ep_n:=\exp\Big\{-\frac{\ln^2n}3\Big\}. 
\end{equation}
So,
\begin{equation}
    P=\tP+O(\ep_n),
\end{equation}
\begin{equation}
    \tP:=P(Z_i^2\le Z_1^2-b\ \forall i\ge2), 
\end{equation}
where
\begin{equation}
    b=na(1+\de_n)=(1+\de_n)c\ln n
    =c\ln n+o(1)\quad\text{and}\quad|\de_n|\le\frac{\sqrt n\,\ln n}n=o(1/\ln n).
\end{equation}
Next, conditioning on $Z_1$, we have
\begin{equation}
    \tP=\int_\R dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1}, 
\end{equation}
where $f$ is the standard normal pdf, $G(t):=P(Z>t)$, and $t_+:=\max(0,t)$. Consider here the substitution
\begin{equation}
    e^{-u}=1-2G(\sqrt{(x^2-b)_+}), 
\end{equation}
with $e^{-u}:=0$ if $x^2-b\le0$; note that $u\ge0$.
For each real $h>0$, the integral of $f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1}=f(x)e^{-(n-1)u}$ over the set of $x>0$ such that $u>h$ is less than $e^{-(n-1)h}$. So, we can choose $h=h_n$ so that
\begin{equation}
h\downarrow0,\quad nh\to\infty,     
\end{equation}
and
\begin{equation}
    \int_{x>0\colon\, u>h} dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1} \le\ep_n
\end{equation}
eventually.
On the other hand, if $u\le h$ and $x>0$, then $u=o(1)$ and hence
\begin{equation}
    \sqrt{(x^2-b)_+}=\sqrt{x^2-b}=z:=G^{-1}\Big(\frac{1-e^{-u}}2\Big)\to\infty,
\end{equation}
\begin{equation}
    x=\sqrt{b+z^2},
\end{equation}
\begin{equation}
    dx=-\frac{du\, z}{\sqrt{b+z^2}}\frac1{f(z)}\frac{e^{-u}}2.
\end{equation}
Next, $G(z)=\frac{1-e^{-u}}2\sim u/2$ and $G(z)=e^{-z^2/(2+o(1))}$, whence
\begin{equation}
    z\sim\sqrt{2\ln\frac1u}, 
\end{equation}
\begin{equation}
    f(z)\sim G(z)z\sim\frac u2\,\sqrt{2\ln\frac1u},
\end{equation}
\begin{equation}
    dx\sim-\frac{du\,\sqrt{2\ln\frac1u}}{\sqrt{b+2\ln\frac1u}}\frac1{\frac u2\,\sqrt{2\ln\frac1u}}\frac{e^{-u}}2
    =-\frac{du}{\sqrt{b+2\ln\frac1u}}\frac{e^{-u}}u.
\end{equation}
Further,
\begin{equation}
    f(x)=e^{-b/2}f(z)\sim e^{-b/2}\frac u2\,\sqrt{2\ln\frac1u}.
\end{equation}
So,
\begin{align*}
&   \int_{x>0\colon\, u\le h} dx\,f(x)(1-2G(\sqrt{(x^2-b)_+})^{n-1} \\ 
    &\sim\int_0^h e^{-b/2}\frac u2\,\sqrt{2\ln\frac1u}\,\frac{du}{\sqrt{b+2\ln\frac1u}}\frac{e^{-u}}u   
    e^{-(n-1)u} \\ 
    &=\frac{e^{-b/2}}2\,\int_0^h du\,\frac{\sqrt{2\ln\frac1u}}{\sqrt{b+2\ln\frac1u}}
    e^{-nu} \\ 
    &=\frac{e^{-b/2}}{2n}\,\int_0^{nh} dy\, e^{-y}\,\frac1{\sqrt{1+\frac b{2\ln\frac ny}}}
     \\ 
    &=\frac{e^{-(c\ln n+o(1))/2}}{2n}\,\int_0^{nh} dy\, e^{-y}\,\frac1{\sqrt{1+\frac{c\ln n+o(1)}{2\ln\frac ny}}}
     \\ 
    &\sim \frac{n^{-1-c/2}}2\,\frac1{\sqrt{1+c/2}}       
     \end{align*}
by dominated convergence.
Collecting all the pieces and noting that the integrand is even in $x$, we conclude that for $a=\frac{c\ln n}n$
\begin{equation}
    P\sim \frac{n^{-1-c/2}}{\sqrt{1+c/2}}
\end{equation}
uniformly over all nonnegative $c$ in any bounded interval.
